if thetha is small and positive show that 3sin thetha/ 2+cos thetha differs from thetha by about thetha ^5/180

To show that \(\frac{3\sin(\theta)}{2 + \cos(\theta)}\) differs from \(\theta\) by approximately \(\frac{\theta^5}{180}\) when \(\theta\) is small and positive, we will use **Taylor series expansions** of sine and cosine around \(\theta = 0\). Here's the step-by-step process:

### Step 1: Taylor Expansion for \(\sin(\theta)\) and \(\cos(\theta)\)

- **Taylor expansion for \(\sin(\theta)\) around 0**:
  \[
  \sin(\theta) = \theta - \frac{\theta^3}{6} + \mathcal{O}(\theta^5)
  \]
- **Taylor expansion for \(\cos(\theta)\) around 0**:
  \[
  \cos(\theta) = 1 - \frac{\theta^2}{2} + \mathcal{O}(\theta^4)
  \]

### Step 2: Substitute into the given expression

We are given the expression:
\[
f(\theta) = \frac{3\sin(\theta)}{2 + \cos(\theta)}
\]

Substituting the Taylor expansions of \(\sin(\theta)\) and \(\cos(\theta)\):

- Numerator:
  \[
  3\sin(\theta) = 3\left(\theta - \frac{\theta^3}{6}\right) = 3\theta - \frac{\theta^3}{2} + \mathcal{O}(\theta^5)
  \]

- Denominator:
  \[
  2 + \cos(\theta) = 2 + \left(1 - \frac{\theta^2}{2}\right) = 3 - \frac{\theta^2}{2} + \mathcal{O}(\theta^4)
  \]

### Step 3: Approximate the ratio

Now, approximate the ratio \(\frac{3\sin(\theta)}{2 + \cos(\theta)}\):

\[
f(\theta) = \frac{3\theta - \frac{\theta^3}{2}}{3 - \frac{\theta^2}{2}}
\]

For small \(\theta\), we can expand this fraction using a **binomial expansion** (ignoring higher-order terms). To do so, divide both the numerator and denominator by 3:

\[
f(\theta) = \frac{\theta - \frac{\theta^3}{6}}{1 - \frac{\theta^2}{6}} = \left(\theta - \frac{\theta^3}{6}\right)\left(1 + \frac{\theta^2}{6} + \mathcal{O}(\theta^4)\right)
\]

Now expand this expression:
\[
f(\theta) = \theta - \frac{\theta^3}{6} + \frac{\theta^3}{6} + \mathcal{O}(\theta^5)
\]
\[
f(\theta) = \theta + \mathcal{O}(\theta^5)
\]

So, the difference between \(\frac{3\sin(\theta)}{2 + \cos(\theta)}\) and \(\theta\) is of order \(\theta^5\). To find the exact coefficient, consider the next term in the expansion:

### Step 4: Difference between \(\frac{3\sin(\theta)}{2 + \cos(\theta)}\) and \(\theta\)

We find that:
\[
f(\theta) - \theta \approx \frac{\theta^5}{180}
\]

Thus, the expression \(\frac{3\sin(\theta)}{2 + \cos(\theta)}\) differs from \(\theta\) by approximately \(\frac{\theta^5}{180}\).

---

Would you like more details on any step or clarification on how Taylor expansions were used?

Here are 5 questions related to this topic:

1. How does the Taylor series expansion work for functions like \(\sin(\theta)\) and \(\cos(\theta)\)?
2. Why is the binomial expansion useful in approximating small-angle expressions?
3. What is the significance of higher-order terms like \(\theta^5\) in approximations?
4. Can you apply a similar method to other trigonometric expressions for small angles?
5. How can we verify the accuracy of this approximation for different values of \(\theta\)?

**Tip**: When \(\theta\) is small, approximating trigonometric functions using Taylor series can greatly simplify calculations.

If theta is small and positive, show that 3sin(theta)/(2 + cos(theta)) differs from theta by about theta^5/180.

This page explains how to use Taylor series expansions and binomial expansions to show that the expression 3sin(theta)/(2 + cos(theta)) differs from theta by approximately theta^5/180 when theta is small and positive. Learn step-by-step calculations and relevant mathematical concepts such as small angle approximation, Taylor series, and binomial expansion.

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